Solve the following quatratic equation :
`a^(2)b^(2)x^(2)+b^(2)x-a^(2)x-1=0`

Options

A. `(-1)/(a)and(1)/(b)`
B. `(-1)/(a^(2))and(1)/(b^(2))`
C. `(1)/(a^(2))and(1)/(b^(2))`
D. `(-1)/(a^(2))and(-1)/(b^(2))`

Explanation

To solve the quadratic equation \( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ a^2b^2x^2 + b^2x - a^2x - 1 = 0 \]
We can rearrange the terms involving \( x \):
\[ a^2b^2x^2 + (b^2 - a^2)x - 1 = 0 \]

### Step 2: Identifying Coefficients
From the rearranged equation, we can identify the coefficients:
- \( A = a^2b^2 \)
- \( B = b^2 - a^2 \)
- \( C = -1 \)

### Step 3: Applying the Quadratic Formula
We will use the quadratic formula to find the values of \( x \):
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the coefficients:
\[ x = \frac{-(b^2 - a^2) \pm \sqrt{(b^2 - a^2)^2 - 4(a^2b^2)(-1)}}{2(a^2b^2)} \]

### Step 4: Simplifying the Discriminant
Calculating the discriminant:
\[ B^2 - 4AC = (b^2 - a^2)^2 + 4a^2b^2 \]
Expanding \( (b^2 - a^2)^2 \):
\[ (b^2 - a^2)^2 = b^4 - 2a^2b^2 + a^4 \]
Thus,
\[ B^2 - 4AC = b^4 - 2a^2b^2 + a^4 + 4a^2b^2 = b^4 + 2a^2b^2 + a^4 = (b^2 + a^2)^2 \]

### Step 5: Substituting Back into the Formula
Now substituting back into the quadratic formula:
\[ x = \frac{-(b^2 - a^2) \pm (b^2 + a^2)}{2a^2b^2} \]

### Step 6: Finding the Roots
We will consider both cases for the \( \pm \):

1. **Case 1**: Using \( + \):
\[ x = \frac{-(b^2 - a^2) + (b^2 + a^2)}{2a^2b^2} = \frac{2a^2}{2a^2b^2} = \frac{1}{b^2} \]

2. **Case 2**: Using \( - \):
\[ x = \frac{-(b^2 - a^2) - (b^2 + a^2)}{2a^2b^2} = \frac{-2b^2}{2a^2b^2} = -\frac{1}{a^2} \]

### Step 7: Final Solutions
Thus, the solutions for the quadratic equation are:
\[ x = \frac{1}{b^2} \quad \text{and} \quad x = -\frac{1}{a^2} \]